Question

A rescue helicopter lifts a 94.8-kg person straight up by means of a cable. The person has an upward acceleration of 0.581 m/s2 and is lifted from rest through a distance of 13.0 m. (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person's weight? (d) Use the work-energy theorem and find the final speed of the person.

Answer 1

here,

mass of person , m = 94.8 kg

accelration , a = 0.581 m/s^2

s = 13 m

a)

the tension in the cable , T = m * ( g + a)

T = 94.8 * ( 9.81 + 0.581) N

T = 985.1 N

b)

the work done by tension force , Wt = T * s

Wt = 985.1 * 13 J = 1.28 * 10^4 J

c)

the work done by person's weight , Wg = m * g * s * cos(180)

Wg = - 94.8 * 9.81 * 13 J

Wg = - 12089.844 J = - 1.21 * 10^4 J

d)

let the speed of person be v

using work-energy theorm

the net work done = kinetic energy gained

Wg + Wt = 0.5 * m * v^2

1.28 * 10^4 - 1.21 * 10^4 = 0.5 * 94.8 * v^2

solving for v

v = 3.89 m/s